Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and (b) Planck's constant.
`{:(lambda (nm),500,450,400),(v xx 10^(-5)(cm s^(-1)), 2.55 , 4.35, 5.20):}`

Explanation

Suppose threshold wavelength `= lamda_(0) nm = lamda_(0) xx 10^(-9)m`
Then `h (v - v_(0)) = (1)/(2) mv^(2) or hc ((1)/(lamda) - (1)/(lamda_(0))) = (1)/(2) mv^(2)`
Substituting the given results of the three experiments, we get
`(hc)/(10^(-9)) ((1)/(500) - (1)/(lamda_(0))) = (1)/(2) m (2.55 xx 10^(6))^(2)`....(i)
`(hc)/(10^(-9)) ((1)/(450) - (1)/(lamda_(0))) = (1)/(2) m (4.35 xx 10^(6))^(2)`...(ii)
`(hc)/(10^(-9)) ((1)/(400) - (1)/(lamda_(0))) = (1)/(2) m (5.20 xx 10^(6))^(2)`
Dividing eqn. by eqn. (i), we get `(lamda_(0) - 450)/(450 lamda_(0)) xx (500 lamda_(0))/(lamda_(0) - 500) =((4.35)/(2.55))^(2)`
or `(lamda_(0) - 450)/(lamda_(0) - 500) xx (450)/(500) = ((4.35)/(2.55))^(2) = 2.619 or lamda_(0) - 450 = 2.619 lamda_(0) - 1309.5`
or `1.619 lamda_(0) = 859.5 :. lamda_(0) = 531nm`
Substituting this value in eqn. (iii), we get
`(h xx (3 xx 10^(8)))/(10^(-9)) ((1)/(400) - (1)/(531)) = (1)/(2) (9.11 xx 10^(-31)) (5.20 xx 10^(6))^(2) or h = 6.66 xx 10^(-34) Js`